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3 years ago

What is the percent composition by mass of oxygen in

Chemistry

1 answer:

KiRa [710]3 years ago

4 0

What is the percent composition by mass of oxygen in magnesium oxide, MgO?

Answer: 39.7 percent

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In liquids, the attractive intermolecular forces are ________. In liquids, the attractive intermolecular forces are ________. st

Fantom [35]

Answer:

strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other.

Explanation:

In liquids, the attractive intermolecular forces are <u>strong enough to hold molecules relatively close together but not strong enough to keep molecules from moving past each other</u>.

Intermolecular forces are the forces of repulsion or attraction.

Intermolecular forces lie between atoms, molecules, or ions. Intramolecular forces are strong in comparison to these forces.

<u />

5 0

3 years ago

1. The pressure of a gas is 100.0 kPa and its volume is 500.0 ml. If the volume increases to 1,000.0 ml, what is the new pressur

marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $P_{1} = 100\,kPa$ , $V_{1} = 500\,mL$ and $V_{2} = 1000\,mL$ , then the new pressure of the gas is:

$P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)$

$P_{2} = 500\,kPa$

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $V_{1} = 3.60\,L$ , $P_{1} = 10\,kPa$ and $P_{2} = 25\,kPa$ then the new volume of the gas is:

$V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)$

$V_{2} = 1.44\,L$

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

$\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}$ (1)

Where:

$P_{1}, P_{2}$ - Initial and final pressure, measured in kPa.

$V_{1}, V_{2}$ - Initial and final pressure, measured in mililiters.

If we know that $\frac{P_{2}}{P_{1}} = 3$ , then the volume ratio is:

$\frac{V_{1}}{V_{2}} = 3$

$\frac{V_{2}}{V_{1}} = \frac{1}{3}$

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

8 0

3 years ago

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

Anastasy [175]

The dissociation equation will be

NH4OH ---> NH4+ + OH-

Initial 0.006 0 0

Change -0.006 X 0.053 +0.006 X 0.053 -0.006 X 0.053

Equlibrium 0.006 -0.006 X 0.053 0.006 X 0.053 0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

7 0

3 years ago

Use Avogadro's number, 6.02E23, to calculate the number

Zepler [3.9K]

Answer:

<h2>2.408 × 10²¹ is the correct answer!!</h2>

8 0

3 years ago

One liter of air contains about 0.21 L of oxygen. When filled, the human lungs hold about 6.0 L of air. How much oxygen is in th

masya89 [10]

Since we are told that 1L of air contains 0.21L of oxygen, you can use the conversion (0.21L O₂)/(1L air). That means that you can just multiply 6.0L by 0.21L to get 1.26L of O₂.
that means that the lungs can hold about 1.26L of oxygen.
I hope this helps. Let me know if anything is unclear.

4 0

3 years ago